Bipartite Graph Check in DSA C - Time & Space Complexity

Time Complexity: Bipartite Graph Check

O(n^2)

Understanding Time Complexity

We want to know how the time needed to check if a graph is bipartite changes as the graph grows.

Specifically, how does the number of nodes (n) affect the work done? (Note: In adjacency matrix representation, edges do not impact asymptotic time.)

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Check if graph is bipartite using BFS
bool isBipartite(int graph[][MAX], int n) {
    int color[n];
    for (int i = 0; i < n; i++) color[i] = -1;
    
    for (int start = 0; start < n; start++) {
        if (color[start] == -1) {
            color[start] = 0;
            Queue q;
            q.push(start);
            while (!q.empty()) {
                int u = q.front(); q.pop();
                for (int v = 0; v < n; v++) {
                    if (graph[u][v] && color[v] == -1) {
                        color[v] = 1 - color[u];
                        q.push(v);
                    } else if (graph[u][v] && color[v] == color[u]) {
                        return false;
                    }
                }
            }
        }
    }
    return true;
}
    

This code checks if a graph is bipartite by coloring nodes using breadth-first search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: BFS dequeues each node once and scans all n possible neighbors (matrix row).
• How many times: n nodes dequeued × n neighbor checks each = O(n²).

How Execution Grows With Input

As the number of nodes (n) grows, the work grows quadratically with n. Edges do not affect asymptotic time (full matrix rows scanned).

Input Size (n)	Approx. Operations
10 nodes	About 100 checks (n²)
100 nodes	About 10,000 checks
1000 nodes	About 1,000,000 checks

Pattern observation: The operations grow quadratically with the number of nodes.

Final Time Complexity

Time Complexity: O(n²)

This means the time needed grows roughly in proportion to n squared due to adjacency matrix neighbor scans.

Common Mistake

[X] Wrong: "The time depends only on the number of nodes like O(n), ignoring neighbor checks."

[OK] Correct: Each node's neighbor check scans the full row of n entries in the matrix.

Interview Connect

Understanding this time complexity helps you explain how graph algorithms scale, a key skill in many coding interviews.

Self-Check

"What if the graph is represented using adjacency lists instead of a matrix? How would the time complexity change?"

Practice

1. What does it mean for a graph to be bipartite?

easy

A. It contains at least one cycle

B. Its vertices can be divided into two sets with no edges inside the same set

C. It has an equal number of vertices and edges

D. All vertices have the same degree

Solution

1. Step 1: Understand bipartite graph definition

   A bipartite graph is one where vertices can be split into two groups so that no edge connects vertices within the same group.

2. Step 2: Compare options with definition

   Only Its vertices can be divided into two sets with no edges inside the same set matches this definition exactly. Other options describe unrelated properties.

3. Final Answer:

   Its vertices can be divided into two sets with no edges inside the same set -> Option B

4. Quick Check:

   Bipartite graph = two sets, no same-set edges [OK]

Hint: Bipartite means two groups with no internal edges [OK]

Common Mistakes:

• Confusing bipartite with equal vertices and edges
• Thinking bipartite means no cycles
• Assuming all vertices have same degree

2. Which of the following C code snippets correctly initializes a graph adjacency list for bipartite check?

easy

A. int graph[5][5] = {{0}};

B. int graph[5]; for(int i=0; i<5; i++) graph[i] = NULL;

C. int graph[][] = {0};

D. int* graph[5]; for(int i=0; i<5; i++) graph[i] = malloc(sizeof(int)*5); memset(graph[i], 0, sizeof(int)*5);

Solution

1. Step 1: Understand adjacency list initialization in C

   Adjacency list uses array of pointers to arrays or linked lists. int* graph[5]; for(int i=0; i<5; i++) graph[i] = malloc(sizeof(int)*5); memset(graph[i], 0, sizeof(int)*5); correctly allocates memory and initializes it.

2. Step 2: Check other options for correctness

   int graph[5][5] = {{0}}; is a fixed 2D array but only zero-initialized; B tries to assign NULL to int array elements (invalid); C has incomplete array declaration.

3. Final Answer:

   int* graph[5]; for(int i=0; i<5; i++) graph[i] = malloc(sizeof(int)*5); memset(graph[i], 0, sizeof(int)*5); -> Option D

4. Quick Check:

   Proper adjacency list init = malloc + memset [OK]

Hint: Use malloc and memset to init adjacency lists in C [OK]

Common Mistakes:

• Trying to assign NULL to int array elements
• Using incomplete array declarations
• Not allocating memory for adjacency lists

3. Given the following graph edges: 0-1, 1-2, 2-3, 3-0, what will the bipartite check function return?

medium

A. Runtime error due to cycle

B. false (graph is not bipartite)

C. true (graph is bipartite)

D. Indeterminate result

Solution

1. Step 1: Analyze the graph structure

   The edges form a cycle of length 4: 0-1-2-3-0. Even length cycles can be bipartite, so this graph is bipartite.

2. Step 2: Check bipartite condition on cycle

   Since the cycle length is even, the graph can be colored with two colors without conflict.

3. Final Answer:

   true (graph is bipartite) -> Option C

4. Quick Check:

   Even cycle = bipartite [OK]

Hint: Even length cycles are bipartite, odd length are not [OK]

Common Mistakes:

• Assuming all cycles are not bipartite
• Confusing cycle length parity
• Thinking runtime error occurs due to cycles

4. Identify the bug in this bipartite check code snippet:

bool isBipartite(int graph[][3], int V) {
  int color[V];
  for(int i=0; i

medium

A. Function does not check disconnected components

B. Color array is not initialized to -1

C. Queue is not initialized properly

D. Incorrect color assignment logic

Solution

1. Step 1: Understand the code logic

   The code colors nodes starting from node 0 and uses BFS to assign colors to neighbors.

2. Step 2: Identify missing checks

   The code only starts BFS from node 0 and does not check other disconnected components, which may cause incorrect results if graph is disconnected.

3. Step 3: Verify other options

   Queue is properly initialized; color array is initialized to -1; color assignment logic is correct.

4. Final Answer:

   Function does not check disconnected components -> Option A

5. Quick Check:

   Check all components for bipartite [OK]

Hint: Always check all graph components in bipartite check [OK]

Common Mistakes:

• Ignoring disconnected graph parts
• Assuming single BFS covers entire graph
• Misunderstanding color assignment

5. You have a graph with 6 vertices and edges: 0-1, 0-3, 1-2, 2-3, 4-5. Which statement is true about its bipartite nature?

hard

A. The graph is bipartite because all components are bipartite

B. The graph is not bipartite because of the cycle 0-1-2-3-0

C. The graph is bipartite only if vertex 4 is connected to vertex 2

D. The graph is not bipartite because vertices 4 and 5 form an odd cycle

Solution

1. Step 1: Analyze graph components

   Vertices 0-1-2-3 form a cycle 0-1-2-3-0 of length 4 (even), which is bipartite. Vertices 4-5 form a separate edge, also bipartite.

2. Step 2: Check bipartite condition for entire graph

   Since all components are bipartite, the entire graph is bipartite.

3. Step 3: Evaluate other options

   The graph is not bipartite because of the cycle 0-1-2-3-0 is wrong because even cycles are bipartite. The graph is bipartite only if vertex 4 is connected to vertex 2 is irrelevant. The graph is not bipartite because vertices 4 and 5 form an odd cycle is wrong because 4-5 is just an edge, no cycle.

4. Final Answer:

   The graph is bipartite because all components are bipartite -> Option A

5. Quick Check:

   All components bipartite = graph bipartite [OK]

Hint: Check each component's bipartite status separately [OK]

Common Mistakes:

• Assuming any cycle means not bipartite
• Ignoring disconnected components
• Confusing edges with cycles